haskell - 为什么我们必须使用 state monad 而不是直接传递 state？

Question

有人可以举一个简单的例子，其中 state monad 比直接传递 state 更好吗？

bar1 (Foo x) = Foo (x + 1)

对比

bar2 :: State Foo Foo
bar2 = do
  modify (\(Foo x) -> Foo (x + 1))
  get

Answer 1

状态传递通常是乏味的、容易出错的并且阻碍重构。例如，尝试按后序标记二叉树或玫瑰树:

data RoseTree a = Node a [RoseTree a] deriving (Show)

postLabel :: RoseTree a -> RoseTree Int
postLabel = fst . go 0 where
  go i (Node _ ts) = (Node i' ts', i' + 1) where

    (ts', i') = gots i ts

    gots i []     = ([], i)
    gots i (t:ts) = (t':ts', i'') where
      (t', i')   = go i t
      (ts', i'') = gots i' ts

在这里，我必须以正确的顺序手动标记状态，传递正确的状态，并且必须确保标签和子节点在结果中的顺序正确(请注意，天真使用 foldr 或 foldl 为 child 节点可能很容易导致不正确的行为)。

另外，如果我尝试将代码更改为预购，我必须进行容易出错的更改:

preLabel :: RoseTree a -> RoseTree Int
preLabel = fst . go 0 where
  go i (Node _ ts) = (Node i ts', i') where -- first change

    (ts', i') = gots (i + 1) ts -- second change

    gots i []     = ([], i)
    gots i (t:ts) = (t':ts', i'') where
      (t', i')   = go i t
      (ts', i'') = gots i' ts

例子:

branch = Node ()
nil  = branch []
tree = branch [branch [nil, nil], nil]
preLabel tree == Node 0 [Node 1 [Node 2 [],Node 3 []],Node 4 []]
postLabel tree == Node 4 [Node 2 [Node 0 [],Node 1 []],Node 3 []]

对比状态单子(monad)解决方案:

import Control.Monad.State
import Control.Applicative

postLabel' :: RoseTree a -> RoseTree Int
postLabel' = (`evalState` 0) . go where
  go (Node _ ts) = do
    ts' <- traverse go ts
    i   <- get <* modify (+1)
    pure (Node i ts')

preLabel' :: RoseTree a -> RoseTree Int
preLabel' = (`evalState` 0) . go where
  go (Node _ ts) = do
    i   <- get <* modify (+1)
    ts' <- traverse go ts
    pure (Node i ts')

这段代码不仅更简洁、更容易正确编写，而且导致订单前或订单后标签的逻辑更加透明。

PS:奖金适用风格:

postLabel' :: RoseTree a -> RoseTree Int
postLabel' = (`evalState` 0) . go where
  go (Node _ ts) =
    flip Node <$> traverse go ts <*> (get <* modify (+1))

preLabel' :: RoseTree a -> RoseTree Int
preLabel' = (`evalState` 0) . go where
  go (Node _ ts) =
    Node <$> (get <* modify (+1)) <*> traverse go ts